24. Balkanska matematiqka olimpijada


 Augustine Wiggins
 8 months ago
 Views:
Transcription
1 4. Balkanska matematika olimpijada Rodos, Gka 8. apil U konveksnom etvoouglu ABCD vaжi AB = BC = CD, dijagonale AC i BD su azliite duжine i seku se u taki E. Dokazati da je AE = DE ako i samo ako je BAD+ ADC = 10. (Albanija). Na i sve funkcije f : R R takve da za sve ealne bojeve x, y vaжi f(f(x) + y) = f(f(x) y) + 4f(x)y. (Bugaska) 3. Na i sve piodne bojeve n za koje postoji pemutacija σ bojeva 1,,..., n takva da je boj σ(1) + σ() + + σ(n) acionalan. (Sbija) 4. Dat je ceo boj n 3. Neka su C 1, C i C 3 ganice ti konveksna ntougla u avni takva da je pesek svake dve od ǌih konaan skup taaka. Na i najve i mogu i boj taaka skupa C 1 C C 3. (Tuska)
2 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas, 8 Apil 007 Poblem 1. Let ABCD be a convex uadilateal with AB = BC = CD, AC BD and let E be the intesection point of its diagonals. Pove that AE = DE if and only if BAD + ADC = 10. Poblem. Find all functions f : such that ( ( ) ) ( ( ) ) ( ) f f x + y = f f x y + 4 f x y, fo any xy,. Poblem 3. Find all positive integes n such that thee is a pemutation σ of the set {1,,,n} fo which σ(1) + σ() + + σ( n) is a ational numbe. Note: A pemutation of the set {1,,,n} is a onetoone function of this set to itself. Poblem 4. Fo a given positive intege n >, let C1, C, C 3 be the boundaies of thee convex ngons in the plane such that C C, C C, C C ae finite. Find the maximum numbe of points of the set C C C 3. 1 Time allowed 4 hous and 30 minutes Each poblem is woth 10 points.
3
4 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem. Find all functions f : R! R such that f(f(x) + y) = f(f(x) y) + 4f(x)y fo any x; y R: Solution. It is clea that the function f 0 satis es the given condition. y Assume that f 6 0: Choose x 0 such that f(x 0 ) 6= 0 and set ey = 4f(x 0 ) y: Then plugging in to the given elation x 0 fo x and ey fo y we get fo any y = f(f(x 0 ) + ey) f(f(x 0 ) ey): (1) Fo any y 1 ; y, plugging in to (1) y 1 y fo y, we get y 1 y y = f(f(x 0 ) + ^ 1 y ) f(f(x 0 ) y^ 1 y ): In othe wods, fo any y 1 ; y thee exist x 1 (y 1 ; y ) = f(x 0 ) + ]y 1 y ; x (y 1 ; y ) = f(x 0 ) + ]y 1 y such that y 1 y = f(x 1 ) f(x ), i.e. such that f(x 1 ) y 1 = f(x ) y : () On the othe hand, eplacing y by f(x) y in the given condition gives f(f(x) y) = f(y) + 4f(x)(f(x) y); i.e. f(y) y = f(f(x) y) ((f(x) y) : (3) Now if fo any two y 1 ; y, we plug in to (3), x 1 (y 1 ; y ) and x (y 1 ; y ) espectively, we get f(y 1 ) y 1 = f(f(x 1 ) y 1 ) ((f(x 1 ) y 1 ) and f(y ) y = f(f(x ) y ) ((f(x ) y ) Then by () we get f(y 1 ) y1 = f(y ) y: Since this happens fo any two y 1 ; y, we conclude that f(x) x = constant fo all x, thus f(x) = x + c; c R. It is easy to check that such a function satis es the condition of the poblem. 1
5 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 3. Find all positive s integes n such that thee is a pemutation of the set f1; ; :::; ng; fo which (1) + () + ::: + p (n) is a ational numbe. Note: A pemutation of the set f1; ; :::; ng is a onetoone function of this set to itself. s Solution. Fo some n N, let (1) + () + ::: + p (n) = 1 Q. Suaing both sides of the euation we get that () + (3) + ::: + p (n) is s also ational. s Using the same easoning ecusively, we get that fo evey k f1; : : : ; ng, (k) + (k + 1) + ::: + p (n) is ational as well. Knowing that the suae oot of a positive intege is eithe intege o iational, we have that p s (n) is intege. Similaly, we get that (k) + (k + 1) + ::: + p (n), fo evey k f1; : : : ; ng, is intege. Note that s fo k = 1 we get 1 N. We de ne a k as a k = n + n + ::: + p n, fo all k 1. It is easy to {z } k pove by induction that a k < p s n + 1, fo evey k 1. Theefoe, we have (1) + () + ::: + p (n) < a n < p n + 1, implying 1 < p n + 1. Let ` be the positive intege that satis es ` n < (`+1). Fo some i; 1 i n, we have (i) = `. We distinguish two cases: Fist case: i 6= n: Then we have ` < s ` + (i + 1) + ::: + p (n) < p n+1 < `+, implying 1
6 s ` + (i + 1) + ` + 1 = ::: + p (n) = ` + 1. But then it follows that giving ` < 1. A contadiction. Second case: i = n: (i + 1) + ::: + p (n) < p n + 1 < ` + ; Fo ` > 1, ` 1 belongs to the set f(1); :::; (n 1)g. Let j < n be such that (j) = ` 1. Similaly to the st case, we have implying ` < s ` 1 + ` 1 + ` + = (j + 1) + ::: + p` < p n + 1 < ` + ; p (j + 1) + ::: + p` = ` + 1, and (j + 1) + ::: + p` < p n + 1 < ` + ; a contadiction. If ` = 1, then n f1; ; 3g. Checking though all the possibilities, it is easy to see that fo n = 1 and n = 3 thee exist pemutations that satisfy the initial condition. Namely, fo n = 1 we have p 1 = 1, and fo n = 3, we have + p 3 + p 1 =. Fo n = thee is no such pemutation.
7 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 4. Fo a given positive intege n >, let C 1 ; C ; C 3 be the boundaies of thee convex ngons in the plane such that the sets C 1 \ C, C \ C 3, C 3 \ C 1 ae nite. Find the maximum numbe of points of the set C 1 \ C \ C 3. Solution 1: Let us st obseve that, if a line intesects a convex n gon at nitely many points, then the numbe of such points is at most. Theefoe any two of the n gons may intesect in at most n points. Choose two of the n gons, C 1 ; C, and say that thei intesection points ae p 1 ; p ; : : : ; p k. Thus k n. Say that the union of the set of vetices of C 1 and C is f 1 ; ; : : : ; n g. We note that it is possible to have i = j fo some i 6= j. We will de ne a onetoone function f fom fp 1 ; p ; : : : ; p k g to f 1 ; ; : : : ; n g as follows. Fist of all, oient all n gons in the clockwise diection. Thus, if one taveses an n gon accoding to this oientation, the inteio is on the ight and the exteio is on the left. Fo evey p i, thee exist pecisely two line segments (of nonzeo length) which ae subsets of C 1 o C, say [ j ; p i ] on C 1 and [ k ; p i ] on C, such that one can appoach to p i via these line segments in the clockwise diection. Suppose, that the two vectos p i j and p i k, in this ode, fom a ight handed coodinate system. Then none of the points on [ j ; p i ] can be on o in the inteio of C, since fo any point on o in the inteio of C, the vectos p i and p i k ae eithe positive multiples of each othe, o fom a left handed coodinate system. In this case we set f(p i ) = j. Othewise we set f(p i ) = k. In both cases, the agument above shows that thee ae no othe intesection points between f(p i ) and p i, in the clockwise diection. Let us now show that f is 1 1. If f(p i ) = f(p l ) = and say (without loss of geneality) belong to C 1, then the st intesection point encounteed when one stats fom and taveses C 1 in the clockwise diection has to be both p i and p l, hence p i = p l. Now let us estimate the numbe of p i s that can be contained by the thid polygon C 3. Each edge of C 3 contains exactly 0; 1 o of the p i s. Suppose that a given edge of C 3 contains of the p i s, say p 1 and p. Since C 1 and C ae convex and thei intesection with C 3 is geneic, they should have vetices between (in the clockwise sense) p 1 and p (with this ode), and outside C 3. We claim that at least one of these vetices is not in the set f(c 1 \ C \ C 3 ). Let 1 C 1 and C espectively be between (in the clockwise sense) p 1 and p (with this ode). If 1 is on o in the inteio of C (o is on o in the inteio of C 1 ), then 1 (o ) is not in the image of f, since ecall that f(p) fo any p C 1 \ C (thus fo any p C 1 \ C \ C 3 ) is a point of one of C 1 ; C not on o in the inteio of the othe. So the claim is established in this case. The emaining case is to assume that none of the vetices of any of C 1 ; C that lie outside C 3 (and between (in the clockwise sense) p 1 and p 1
8 (with this ode)) also lies in the inteio of the othe of C 1 ; C. In this case clealy the polygons C 1 and C must meet at some point between (in the clockwise sense) p 1 and p (with this ode). Say p 3 is the closest to p 1 such point. Then clealy f(p 3 ) is a vetex of one of C 1 ; C between (in the clockwise sense) p 1 and p (with this ode). These pats of C 1 ; C though, lie outside C 3 ; the inteio of C 3 lie on the othe side of the line p 1 p. Thus f(p 3 ) is not in f(c 1 \ C \ C 3 ) and the claim is established in all cases. Fo the side a of C 3 containing p 1 ; p let us call (a) a vetex as the one in the claim we just poved. It is easy to see that fo distinct sides a; b of C 3 that contain two of the p s, the points a ; b ae distinct. Indeed, let a contain p 1 ; p and b contain p 3 ; p 4 among the p s. If one of a ; b belongs to one of C 1 ; C and the othe does not belong to it, we ae okay. If both a ; b belong to say C 1, then in a clockwise tou aound C 1 stating at p 1, we meet p 1 ; p ; p 3 ; p 4 in this ode. If not, say the ode is p 1 ; p 3 ; p ; p 4. Then the segments p 1 p, p 3 ; p 4 intesect at an inteio point, since C 1 is a convex polygon. But then the sides a; b of C 3 have a common inteio point, a contadiction. So the coect ode is p 1 ; p ; p 3 ; p 4. But we know that adding (a); (b) in this tou the coect ode is p 1 ; (a); p ; p 3 ; (b); p 4. Thus (a); (b) ae distinct as claimed. Now if x of the edges of C 3 contain 1 of the p i s and y of them contain of the p i s, then x + y numbe of sides of C 3, i.e. x + y n. The numbe of points in C 1 \ C \ C 3 is x + y. Since f is injective, x + y is also the numbe of s in f(c 1 \ C \ C 3 ). Also, by the agument in the pevious paagaph, we see that fo evey distinct edge of C 3 containing points we can assign a coesponding distinct i outside the image of f(c 1 \ C \ C 3 ). Theefoe x is less o eual to the numbe of s that do not belong in f(c 1 \ C \ C 3 ). So (x + y) + y is at most as much as the numbe of s. I.e. x + 3y n. Adding this with x + y n and dividing by, and also taking into account that x + y is an intege x + y b 3n c Let us now show that this is the best uppe bound fo evey n 3. One way (among many) to constuct an example is as follows: Constuct two egula n gons C 1 ; C with the same cente, such that thei intesection points fom a egula n gon. Call the vetices p 1 ; p ; : : : ; p n in a cyclic ode. Let the cicumcicle of this n gon be C. Then let the n gon bounded by the lines p 1 p 3 ; p 5 p 7 ; p 9 p 11 ; : : : (including p k+1 p 1 in case n is an odd n = k + 1) togethe with the tangent lines to C at p 4 ; p 8 ; p 1 ; : : : be C 3. It can easily be checked that jc 1 \ C \ C 3 j = b 3nc. Solution : Let A and B be two conseutive points of C 1 \ C \ C 3 obseved in the clockwise diection fom a point in the inteio of all thee ngons. Let s look fo each C i its section in the clockwise diection between A and B exluding these points. If some two of these sections both do not contain any vetices of thei coesponding ngons, then the segment AB belongs to both ngons, a contadiction.
9 Thus at least two of these segments have at least one vetex each, and moeovethey do not contain the segment. Tivially, two distinct such vetices exist. Since thee exist jc 1 \ C \ C 3 j many conseutive points points A and B of C 1 \ C \ C 3, thee should exist at least jc 1 \ C \ C 3 j distinct vetices of the thee ngons. Thus jc 1 \ C \ C 3 j 3n i.e. jc 1 \ C \ C 3 j b 3nc since (jc 1 \ C \ C 3 j is an intege as well). Actually we can achieve this uppe bound by the example given in the Solution 1. 3
of the contestants play as Falco, and 1 6
JHMT 05 Algeba Test Solutions 4 Febuay 05. In a Supe Smash Bothes tounament, of the contestants play as Fox, 3 of the contestants play as Falco, and 6 of the contestants play as Peach. Given that thee
More informationCOLLAPSING WALLS THEOREM
COLLAPSING WALLS THEOREM IGOR PAK AND ROM PINCHASI Abstact. Let P R 3 be a pyamid with the base a convex polygon Q. We show that when othe faces ae collapsed (otated aound the edges onto the plane spanned
More informationA proof of the binomial theorem
A poof of the binomial theoem If n is a natual numbe, let n! denote the poduct of the numbes,2,3,,n. So! =, 2! = 2 = 2, 3! = 2 3 = 6, 4! = 2 3 4 = 24 and so on. We also let 0! =. If n is a nonnegative
More informationStanford University CS259Q: Quantum Computing Handout 8 Luca Trevisan October 18, 2012
Stanfod Univesity CS59Q: Quantum Computing Handout 8 Luca Tevisan Octobe 8, 0 Lectue 8 In which we use the quantum Fouie tansfom to solve the peiodfinding poblem. The Peiod Finding Poblem Let f : {0,...,
More informationPrerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,
R Pena Towe, Road No, Contactos Aea, Bistupu, Jamshedpu 8, Tel (657)89, www.penaclasses.com IIT JEE Mathematics Pape II PART III MATHEMATICS SECTION I Single Coect Answe Type This section contains 8 multiple
More informationChapter 3: Theory of Modular Arithmetic 38
Chapte 3: Theoy of Modula Aithmetic 38 Section D Chinese Remainde Theoem By the end of this section you will be able to pove the Chinese Remainde Theoem apply this theoem to solve simultaneous linea conguences
More informationFREE Download Study Package from website: &
.. Linea Combinations: (a) (b) (c) (d) Given a finite set of vectos a b c,,,... then the vecto xa + yb + zc +... is called a linea combination of a, b, c,... fo any x, y, z... R. We have the following
More information15 Solving the Laplace equation by Fourier method
5 Solving the Laplace equation by Fouie method I aleady intoduced two o thee dimensional heat equation, when I deived it, ecall that it taes the fom u t = α 2 u + F, (5.) whee u: [0, ) D R, D R is the
More informationEM Boundary Value Problems
EM Bounday Value Poblems 10/ 9 11/ By Ilekta chistidi & Lee, SeungHyun A. Geneal Desciption : Maxwell Equations & Loentz Foce We want to find the equations of motion of chaged paticles. The way to do
More informationPROBLEM SET #1 SOLUTIONS by Robert A. DiStasio Jr.
POBLM S # SOLUIONS by obet A. DiStasio J. Q. he BonOppenheime appoximation is the standad way of appoximating the gound state of a molecula system. Wite down the conditions that detemine the tonic and
More informationS7: Classical mechanics problem set 2
J. Magoian MT 9, boowing fom J. J. Binney s 6 couse S7: Classical mechanics poblem set. Show that if the Hamiltonian is indepdent of a genealized coodinate q, then the conjugate momentum p is a constant
More informationSo, if we are finding the amount of work done over a nonconservative vector field F r, we do that long ur r b ur =
3.4 Geen s Theoem Geoge Geen: selftaught English scientist, 79384 So, if we ae finding the amount of wok done ove a nonconsevative vecto field F, we do that long u b u 3. method Wok = F d F( () t )
More informationVECTOR MECHANICS FOR ENGINEERS: STATICS
4 Equilibium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Fedinand P. Bee E. Russell Johnston, J. of Rigid Bodies Lectue Notes: J. Walt Ole Texas Tech Univesity Contents Intoduction FeeBody Diagam
More informationAnalysis of simple branching trees with TI92
Analysis of simple banching tees with TI9 Dušan Pagon, Univesity of Maibo, Slovenia Abstact. In the complex plane we stat at the cente of the coodinate system with a vetical segment of the length one
More informationThe geometric construction of Ewald sphere and Bragg condition:
The geometic constuction of Ewald sphee and Bagg condition: The constuction of Ewald sphee must be done such that the Bagg condition is satisfied. This can be done as follows: i) Daw a wave vecto k in
More informationPermutations and Combinations
Pemutations and Combinations Mach 11, 2005 1 Two Counting Pinciples Addition Pinciple Let S 1, S 2,, S m be subsets of a finite set S If S S 1 S 2 S m, then S S 1 + S 2 + + S m Multiplication Pinciple
More informationGeometry Unit 4b  Notes Triangle Relationships
Geomety Unit 4b  Notes Tiangle Relationships This unit is boken into two pats, 4a & 4b. test should be given following each pat. quidistant fom two points the same distance fom one point as fom anothe.
More informationDeterministic vs Nondeterministic Graph Property Testing
Deteministic vs Nondeteministic Gaph Popety Testing Lio Gishboline Asaf Shapia Abstact A gaph popety P is said to be testable if one can check whethe a gaph is close o fa fom satisfying P using few andom
More informationGreen s Identities and Green s Functions
LECTURE 7 Geen s Identities and Geen s Functions Let us ecall The ivegence Theoem in ndimensions Theoem 7 Let F : R n R n be a vecto field ove R n that is of class C on some closed, connected, simply
More informationTHE JEU DE TAQUIN ON THE SHIFTED RIM HOOK TABLEAUX. Jaejin Lee
Koean J. Math. 23 (2015), No. 3, pp. 427 438 http://dx.doi.og/10.11568/kjm.2015.23.3.427 THE JEU DE TAQUIN ON THE SHIFTED RIM HOOK TABLEAUX Jaejin Lee Abstact. The Schensted algoithm fist descibed by Robinson
More informationChapter 1: Introduction to Polar Coordinates
Habeman MTH Section III: ola Coodinates and Comple Numbes Chapte : Intoduction to ola Coodinates We ae all comfotable using ectangula (i.e., Catesian coodinates to descibe points on the plane. Fo eample,
More informationSolution to HW 3, Ma 1a Fall 2016
Solution to HW 3, Ma a Fall 206 Section 2. Execise 2: Let C be a subset of the eal numbes consisting of those eal numbes x having the popety that evey digit in the decimal expansion of x is, 3, 5, o 7.
More informationMATH 415, WEEK 3: ParameterDependence and Bifurcations
MATH 415, WEEK 3: PaameteDependence and Bifucations 1 A Note on Paamete Dependence We should pause to make a bief note about the ole played in the study of dynamical systems by the system s paametes.
More informationThe ErdősHajnal conjecture for rainbow triangles
The EdősHajnal conjectue fo ainbow tiangles Jacob Fox Andey Ginshpun János Pach Abstact We pove that evey 3coloing of the edges of the complete gaph on n vetices without a ainbow tiangle contains a set
More informationLECTURE 12. Special Solutions of Laplace's Equation. 1. Separation of Variables with Respect to Cartesian Coordinates. Suppose.
50 LECTURE 12 Special Solutions of Laplace's Equation 1. Sepaation of Vaiables with Respect to Catesian Coodinates Suppose è12.1è èx; yè =XèxèY èyè is a solution of è12.2è Then we must have è12.3è 2 x
More informationRADIAL POSITIVE SOLUTIONS FOR A NONPOSITONE PROBLEM IN AN ANNULUS
Electonic Jounal of Diffeential Equations, Vol. 04 (04), o. 9, pp. 0. ISS: 07669. UL: http://ejde.math.txstate.edu o http://ejde.math.unt.edu ftp ejde.math.txstate.edu ADIAL POSITIVE SOLUTIOS FO A OPOSITOE
More information15.081J/6.251J Introduction to Mathematical Programming. Lecture 6: The Simplex Method II
15081J/6251J Intoduction to Mathematical Pogamming ectue 6: The Simplex Method II 1 Outline Revised Simplex method Slide 1 The full tableau implementation Anticycling 2 Revised Simplex Initial data: A,
More informationPhysics 107 TUTORIAL ASSIGNMENT #8
Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type
More information3.6 Applied Optimization
.6 Applied Optimization Section.6 Notes Page In this section we will be looking at wod poblems whee it asks us to maimize o minimize something. Fo all the poblems in this section you will be taking the
More informationRelated Rates  the Basics
Related Rates  the Basics In this section we exploe the way we can use deivatives to find the velocity at which things ae changing ove time. Up to now we have been finding the deivative to compae the
More informationA Crash Course in (2 2) Matrices
A Cash Couse in ( ) Matices Seveal weeks woth of matix algeba in an hou (Relax, we will only stuy the simplest case, that of matices) Review topics: What is a matix (pl matices)? A matix is a ectangula
More informationIntroduction Common Divisors. Discrete Mathematics Andrei Bulatov
Intoduction Common Divisos Discete Mathematics Andei Bulatov Discete Mathematics Common Divisos 3 Pevious Lectue Integes Division, popeties of divisibility The division algoithm Repesentation of numbes
More information2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum
2. Electostatics D. Rakhesh Singh Kshetimayum 1 2.1 Intoduction In this chapte, we will study how to find the electostatic fields fo vaious cases? fo symmetic known chage distibution fo unsymmetic known
More informationGalois points on quartic surfaces
J. Math. Soc. Japan Vol. 53, No. 3, 2001 Galois points on quatic sufaces By Hisao Yoshihaa (Received Nov. 29, 1999) (Revised Ma. 30, 2000) Abstact. Let S be a smooth hypesuface in the pojective thee space
More informationAs is natural, our Aerospace Structures will be described in a Euclidean threedimensional space R 3.
Appendix A Vecto Algeba As is natual, ou Aeospace Stuctues will be descibed in a Euclidean theedimensional space R 3. A.1 Vectos A vecto is used to epesent quantities that have both magnitude and diection.
More informationWhen two numbers are written as the product of their prime factors, they are in factored form.
10 1 Study Guide Pages 420 425 Factos Because 3 4 12, we say that 3 and 4 ae factos of 12. In othe wods, factos ae the numbes you multiply to get a poduct. Since 2 6 12, 2 and 6 ae also factos of 12. The
More informationTarget Boards, JEE Main & Advanced (IIT), NEET Physics Gauss Law. H. O. D. Physics, Concept Bokaro Centre P. K. Bharti
Page 1 CONCPT: JB, Nea Jitenda Cinema, City Cente, Bokao www.vidyadishti.og Gauss Law Autho: Panjal K. Bhati (IIT Khaagpu) Mb: 74884484 Taget Boads, J Main & Advanced (IIT), NT 15 Physics Gauss Law Autho:
More informationConnectedness of Ordered Rings of Fractions of C(X) with the mtopology
Filomat 31:18 (2017), 5685 5693 https://doi.og/10.2298/fil1718685s Published by Faculty o Sciences and Mathematics, Univesity o Niš, Sebia Available at: http://www.pm.ni.ac.s/ilomat Connectedness o Odeed
More informationB. Spherical Wave Propagation
11/8/007 Spheical Wave Popagation notes 1/1 B. Spheical Wave Popagation Evey antenna launches a spheical wave, thus its powe density educes as a function of 1, whee is the distance fom the antenna. We
More informationPHYS 705: Classical Mechanics. Small Oscillations
PHYS 705: Classical Mechanics Small Oscillations Fomulation of the Poblem Assumptions: V q  A consevative system with depending on position only  The tansfomation equation defining does not dep on time
More informationOutputSensitive Algorithms for Computing NearestNeighbour Decision Boundaries
OutputSensitive Algoithms fo Computing NeaestNeighbou Decision Boundaies David Bemne 1, Eik Demaine 2, Jeff Eickson 3, John Iacono 4, Stefan Langeman 5, Pat Moin 6, and Godfied Toussaint 7 1 Faculty
More informationRelating Branching Program Size and. Formula Size over the Full Binary Basis. FB Informatik, LS II, Univ. Dortmund, Dortmund, Germany
Relating Banching Pogam Size and omula Size ove the ull Binay Basis Matin Saueho y Ingo Wegene y Ralph Wechne z y B Infomatik, LS II, Univ. Dotmund, 44 Dotmund, Gemany z ankfut, Gemany sauehof/wegene@ls.cs.unidotmund.de
More informationMagnetic Field. Conference 6. Physics 102 General Physics II
Physics 102 Confeence 6 Magnetic Field Confeence 6 Physics 102 Geneal Physics II Monday, Mach 3d, 2014 6.1 Quiz Poblem 6.1 Think about the magnetic field associated with an infinite, cuent caying wie.
More information3.2 Centripetal Acceleration
unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tackandfield event in which an athlete thows a hamme
More informationAdditive Approximation for EdgeDeletion Problems
Additive Appoximation fo EdgeDeletion Poblems Noga Alon Asaf Shapia Benny Sudakov Abstact A gaph popety is monotone if it is closed unde emoval of vetices and edges. In this pape we conside the following
More informationMath Section 4.2 Radians, Arc Length, and Area of a Sector
Math 1330  Section 4. Radians, Ac Length, and Aea of a Secto The wod tigonomety comes fom two Geek oots, tigonon, meaning having thee sides, and mete, meaning measue. We have aleady defined the six basic
More informationEnergy Levels Of Hydrogen Atom Using Ladder Operators. Ava Khamseh Supervisor: Dr. Brian Pendleton The University of Edinburgh August 2011
Enegy Levels Of Hydogen Atom Using Ladde Opeatos Ava Khamseh Supeviso: D. Bian Pendleton The Univesity of Edinbugh August 11 1 Abstact The aim of this pape is to fist use the Schödinge wavefunction methods
More information(A) 2log( tan cot ) [ ], 2 MATHEMATICS. 1. Which of the following is correct?
MATHEMATICS. Which of the following is coect? A L.P.P always has unique solution Evey L.P.P has an optimal solution A L.P.P admits two optimal solutions If a L.P.P admits two optimal solutions then it
More informationMotion in One Dimension
Motion in One Dimension Intoduction: In this lab, you will investigate the motion of a olling cat as it tavels in a staight line. Although this setup may seem ovesimplified, you will soon see that a detailed
More informationWe give improved upper bounds for the number of primitive solutions of the Thue inequality
NUMBER OF SOLUTIONS OF CUBIC THUE INEQUALITIES WITH POSITIVE DISCRIMINANT N SARADHA AND DIVYUM SHARMA Abstact Let F(X, Y) be an ieducible binay cubic fom with intege coefficients and positive disciminant
More informationReduced Implicant Tries
Reduced Implicant Ties Technical Repot SUNYACS0701 Novembe, 2007 Neil V. Muay Depatment of Compute Science Univesity at Albany Albany, NY 12222 email: nvm@cs.albany.edu Eik Rosenthal Depatment of Mathematics
More informationRadian and Degree Measure
CHAT PeCalculus Radian and Degee Measue *Tigonomety comes fom the Geek wod meaning measuement of tiangles. It pimaily dealt with angles and tiangles as it petained to navigation, astonomy, and suveying.
More informationLecture 7: Angular Momentum, Hydrogen Atom
Lectue 7: Angula Momentum, Hydogen Atom Vecto Quantization of Angula Momentum and Nomalization of 3D Rigid Roto wavefunctions Conside l, so L 2 2 2. Thus, we have L 2. Thee ae thee possibilities fo L z
More informationCOMP Parallel Computing SMM (3) OpenMP Case Study: The BarnesHut Nbody Algorithm
COMP 633  Paallel Computing Lectue 8 Septembe 14, 2017 SMM (3) OpenMP Case Study: The BanesHut Nbody Algoithm Topics Case study: the BanesHut algoithm Study an impotant algoithm in scientific computing»
More informationGeneralisations of a FourSquare Theorem. Hiroshi Okumura and John F. Rigby. results on a wooden board and dedicated it to a shrine or a temple.
20 enealisations of a ousquae Theoem Hioshi Okumua and John. Rigby In the 17 th {19 th centuies, Japanese people often wote thei mathematical esults on a wooden boad and dedicated it to a shine o a temple.
More informationMULTILAYER PERCEPTRONS
Last updated: Nov 26, 2012 MULTILAYER PERCEPTRONS Outline 2 Combining Linea Classifies Leaning Paametes Outline 3 Combining Linea Classifies Leaning Paametes Implementing Logical Relations 4 AND and OR
More informationNumerical Integration
MCEN 473/573 Chapte 0 Numeical Integation Fall, 2006 Textbook, 0.4 and 0.5 Isopaametic Fomula Numeical Integation [] e [ ] T k = h B [ D][ B] e B Jdsdt In pactice, the element stiffness is calculated numeically.
More informationNonLinear Dynamics Homework Solutions Week 2
NonLinea Dynamics Homewok Solutions Week Chis Small Mach, 7 Please email me at smach9@evegeen.edu with any questions o concens eguading these solutions. Fo the ececises fom section., we sketch all qualitatively
More informationPhysics 111 Lecture 5 Circular Motion
Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight
More informationNOTE. Some New Bounds for CoverFree Families
Jounal of Combinatoial Theoy, Seies A 90, 224234 (2000) doi:10.1006jcta.1999.3036, available online at http:.idealibay.com on NOTE Some Ne Bounds fo CoveFee Families D. R. Stinson 1 and R. Wei Depatment
More informationBoundary Layers and Singular Perturbation Lectures 16 and 17 Boundary Layers and Singular Perturbation. x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9.
Lectues 16 and 17 Bounday Layes and Singula Petubation A Regula Petubation In some physical poblems, the solution is dependent on a paamete K. When the paamete K is vey small, it is natual to expect that
More informationVectors, Vector Calculus, and Coordinate Systems
! Revised Apil 11, 2017 1:48 PM! 1 Vectos, Vecto Calculus, and Coodinate Systems David Randall Physical laws and coodinate systems Fo the pesent discussion, we define a coodinate system as a tool fo descibing
More informationASTR415: Problem Set #6
ASTR45: Poblem Set #6 Cuan D. Muhlbege Univesity of Mayland (Dated: May 7, 27) Using existing implementations of the leapfog and RungeKutta methods fo solving coupled odinay diffeential equations, seveal
More information(Sample 3) Exam 1  Physics Patel SPRING 1998 FORM CODE  A (solution key at end of exam)
(Sample 3) Exam 1  Physics 202  Patel SPRING 1998 FORM CODE  A (solution key at end of exam) Be sue to fill in you student numbe and FORM lette (A, B, C) on you answe sheet. If you foget to include
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II September 15, 2012 Prof. Alan Guth PROBLEM SET 2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Depatment Physics 8.07: Electomagnetism II Septembe 5, 202 Pof. Alan Guth PROBLEM SET 2 DUE DATE: Monday, Septembe 24, 202. Eithe hand it in at the lectue,
More informationEfficiency Loss in a Network Resource Allocation Game
Efficiency Loss in a Netwok Resouce Allocation Game Ramesh Johai johai@mit.edu) John N. Tsitsiklis jnt@mit.edu) June 11, 2004 Abstact We exploe the popeties of a congestion game whee uses of a congested
More informationSeidel s Trapezoidal Partitioning Algorithm
CS68: Geometic Agoithms Handout #6 Design and Anaysis Oigina Handout #6 Stanfod Univesity Tuesday, 5 Febuay 99 Oigina Lectue #7: 30 Januay 99 Topics: Seide s Tapezoida Patitioning Agoithm Scibe: Michae
More informationPhysics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism
Physics 2020, Sping 2005 Lab 5 page 1 of 8 Lab 5. Magnetism PART I: INTRODUCTION TO MAGNETS This week we will begin wok with magnets and the foces that they poduce. By now you ae an expet on setting up
More informationON SPARSELY SCHEMMEL TOTIENT NUMBERS. Colin Defant 1 Department of Mathematics, University of Florida, Gainesville, Florida
#A8 INTEGERS 5 (205) ON SPARSEL SCHEMMEL TOTIENT NUMBERS Colin Defant Depatment of Mathematics, Univesity of Floida, Gainesville, Floida cdefant@ufl.edu Received: 7/30/4, Revised: 2/23/4, Accepted: 4/26/5,
More information0606 ADDITIONAL MATHEMATICS
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Intenational Geneal Cetificate of Seconday Education MARK SCHEME fo the Octobe/Novembe 011 question pape fo the guidance of teaches 0606 ADDITIONAL MATHEMATICS
More informationCOMPUTATIONS OF ELECTROMAGNETIC FIELDS RADIATED FROM COMPLEX LIGHTNING CHANNELS
Pogess In Electomagnetics Reseach, PIER 73, 93 105, 2007 COMPUTATIONS OF ELECTROMAGNETIC FIELDS RADIATED FROM COMPLEX LIGHTNING CHANNELS T.X. Song, Y.H. Liu, and J.M. Xiong School of Mechanical Engineeing
More information$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer
Physics 8.0 Quiz One Equations Fall 006 F = 1 4" o q 1 q = q q ˆ 3 4" o = E 4" o ˆ = points fom souce q to obseve 1 dq E = # ˆ 4" 0 V "## E "d A = Q inside closed suface o d A points fom inside to V =
More informationME 425: Aerodynamics
ME 5: Aeodynamics D ABM Toufique Hasan Pofesso Depatment of Mechanical Engineeing, BUET Lectue 8 Apil 7 teachebuetacbd/toufiquehasan/ toufiquehasan@mebuetacbd ME5: Aeodynamics (Jan 7) Flow ove a stationay
More information( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.
Homewok #8. hapte 7 Magnetic ields. 6 Eplain how ou would modif Gauss s law if scientists discoveed that single, isolated magnetic poles actuall eisted. Detemine the oncept Gauss law fo magnetism now eads
More information4. Electrodynamic fields
4. Electodynamic fields D. Rakhesh Singh Kshetimayum 1 4.1 Intoduction Electodynamics Faaday s law Maxwell s equations Wave equations Lenz s law Integal fom Diffeential fom Phaso fom Bounday conditions
More informationSPECTRAL SEQUENCES: FRIEND OR FOE?
SPECTRAL SEQUENCES: FRIEND OR FOE? RAVI VAKIL Spectal sequences ae a poweful bookkeeping tool fo poving things involving complicated commutative diagams. They wee intoduced by Leay in the 94 s at the
More informationGauss Law. Physics 231 Lecture 21
Gauss Law Physics 31 Lectue 1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing
More informationThis brief note explains why the MichelLevy colour chart for birefringence looks like this...
This bief note explains why the MichelLevy colou chat fo biefingence looks like this... Theoy of Levy Colou Chat fo Biefingent Mateials Between Cossed Polas Biefingence = n n, the diffeence of the efactive
More informationAn Exact Solution of Navier Stokes Equation
An Exact Solution of Navie Stokes Equation A. Salih Depatment of Aeospace Engineeing Indian Institute of Space Science and Technology, Thiuvananthapuam, Keala, India. July 20 The pincipal difficulty in
More informationAP Physics C: Electricity and Magnetism 2001 Scoring Guidelines
AP Physics C: Electicity and Magnetism 1 Scoing Guidelines The mateials included in these files ae intended fo noncommecial use by AP teaches fo couse and exam pepaation; pemission fo any othe use must
More informationGROWTH ESTIMATES THROUGH SCALING FOR QUASILINEAR PARTIAL DIFFERENTIAL EQUATIONS
Annales Academiæ Scientiaum Fennicæ Mathematica Volumen 32, 2007, 595 599 GROWTH ESTIMATES THROUGH SCALING FOR QUASILINEAR PARTIAL DIFFERENTIAL EQUATIONS Teo Kilpeläinen, Henik Shahgholian and Xiao Zhong
More informationNew lower bounds for the independence number of sparse graphs and hypergraphs
New lowe bounds fo the independence numbe of spase gaphs and hypegaphs Kunal Dutta, Dhuv Mubayi, and C.R. Subamanian May 23, 202 Abstact We obtain new lowe bounds fo the independence numbe of K fee gaphs
More informationSection 8.2 Polar Coordinates
Section 8. Pola Coodinates 467 Section 8. Pola Coodinates The coodinate system we ae most familia with is called the Catesian coodinate system, a ectangula plane divided into fou quadants by the hoizontal
More informationrt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)
Cicula Motion Fom ancient times cicula tajectoies hae occupied a special place in ou model of the Uniese. Although these obits hae been eplaced by the moe geneal elliptical geomety, cicula motion is still
More informationGoodnessoffit for composite hypotheses.
Section 11 Goodnessoffit fo composite hypotheses. Example. Let us conside a Matlab example. Let us geneate 50 obsevations fom N(1, 2): X=nomnd(1,2,50,1); Then, unning a chisquaed goodnessoffit test
More informationFunctions Defined on Fuzzy Real Numbers According to Zadeh s Extension
Intenational Mathematical Foum, 3, 2008, no. 16, 763776 Functions Defined on Fuzzy Real Numbes Accoding to Zadeh s Extension Oma A. AbuAaqob, Nabil T. Shawagfeh and Oma A. AbuGhneim 1 Mathematics Depatment,
More informationDynamic Visualization of Complex Integrals with Cabri II Plus
Dynamic Visualiation of omplex Integals with abi II Plus Sae MIKI Kawaijuu, IES Japan Email: sand_pictue@hotmailcom Abstact: Dynamic visualiation helps us undestand the concepts of mathematics This pape
More informationQualifying Examination Electricity and Magnetism Solutions January 12, 2006
1 Qualifying Examination Electicity and Magnetism Solutions Januay 12, 2006 PROBLEM EA. a. Fist, we conside a unit length of cylinde to find the elationship between the total chage pe unit length λ and
More information1 Notes on Order Statistics
1 Notes on Ode Statistics Fo X a andom vecto in R n with distibution F, and π S n, define X π by and F π by X π (X π(1),..., X π(n) ) F π (x 1,..., x n ) F (x π 1 (1),..., x π 1 (n)); then the distibution
More informationClass #16 Monday, March 20, 2017
D. Pogo Class #16 Monday, Mach 0, 017 D NonCatesian Coodinate Systems A point in space can be specified by thee numbes:, y, and z. O, it can be specified by 3 diffeent numbes:,, and z, whee = cos, y =
More informationn 1 Cov(X,Y)= ( X i X )( Y iy ). N1 i=1 * If variable X and variable Y tend to increase together, then c(x,y) > 0
Covaiance and Peason Coelation Vatanian, SW 540 Both covaiance and coelation indicate the elationship between two (o moe) vaiables. Neithe the covaiance o coelation give the slope between the X and Y vaiable,
More informationTopic 4a Introduction to Root Finding & Bracketing Methods
/8/18 Couse Instucto D. Raymond C. Rumpf Office: A 337 Phone: (915) 747 6958 E Mail: cumpf@utep.edu Topic 4a Intoduction to Root Finding & Backeting Methods EE 4386/531 Computational Methods in EE Outline
More informationConspiracy and Information Flow in the TakeGrant Protection Model
Conspiacy and Infomation Flow in the TakeGant Potection Model Matt Bishop Depatment of Compute Science Univesity of Califonia at Davis Davis, CA 956168562 ABSTRACT The Take Gant Potection Model is a
More informationThe 33 rd Balkan Mathematical Olympiad Tirana, May 7, 2016
Language: English The 33 rd Balkan Mathematical Olympiad Tirana, May 7, 2016 Problem 1. Find all injective functions f : R R such that for every real number x and every positive integer n, n i f(x + i
More informationRadian Measure CHAPTER 5 MODELLING PERIODIC FUNCTIONS
5.4 Radian Measue So fa, ou hae measued angles in degees, with 60 being one eolution aound a cicle. Thee is anothe wa to measue angles called adian measue. With adian measue, the ac length of a cicle is
More informationCALCULUS II Vectors. Paul Dawkins
CALCULUS II Vectos Paul Dawkins Table of Contents Peface... ii Vectos... 3 Intoduction... 3 Vectos The Basics... 4 Vecto Aithmetic... 8 Dot Poduct... 13 Coss Poduct... 21 2007 Paul Dawkins i http://tutoial.math.lama.edu/tems.aspx
More informationPhysics 111 Lecture 5 (Walker: 3.36) Vectors & Vector Math Motion Vectors Sept. 11, 2009
Physics 111 Lectue 5 (Walke: 3.36) Vectos & Vecto Math Motion Vectos Sept. 11, 2009 Quiz Monday  Chap. 2 1 Resolving a vecto into xcomponent & y component: Pola Coodinates Catesian Coodinates x y =
More informationA HartreeFock Example Using Helium
Univesity of Connecticut DigitalCommons@UConn Chemisty Education Mateials Depatment of Chemisty June 6 A HateeFock Example Using Helium Cal W. David Univesity of Connecticut, Cal.David@uconn.edu Follow
More informationSPREADING SPEED AND PROFILE FOR NONLINEAR STEFAN PROBLEMS IN HIGH SPACE DIMENSIONS
SPREADING SPEED AND PROFILE FOR NONLINEAR STEFAN PROBLEMS IN HIGH SPACE DIMENSIONS YIHONG DU, HIROSHI MATSUZAWA AND MAOLIN ZHOU Abstact. We conside nonlinea diffusion poblems of the fom u t = u + f(u)
More informationMath Notes on Kepler s first law 1. r(t) kp(t)
Math 7  Notes on Keple s fist law Planetay motion and Keple s Laws We conside the motion of a single planet about the sun; fo simplicity, we assign coodinates in R 3 so that the position of the sun is
More information